Bayesian inference out of partial information - Dirichlet example Suppose we have two coins $X_1$ and $X_2$. They are possibly biased and correlated coins. The heads probability of each coins is denoted by $p_1$ and $p_2$ which we don't know at the beginning. The objective is to estimate the probabilities of HH, TT, H|H, H|T or so on as well as estimating $p_1$ and $p_2$.
To begin, I assume that $$p=(p_{HH},p_{HT},p_{TH},p_{TT})\sim Dir(a_{HH},a_{HT},a_{TH},a_{TT}).$$
If experiments (tossings of two coins, in this case) run in a perfect condition, it would be easy to make a Bayesian inference out of the experiments: we just update the parameter $a$. 
However, suppose that only one coins, say coin 1, is tossed and the outcome is "H". What would the posterior look like in this case? 
I think one of my three attempts should be correct, but I'm not sure which one should be.

Answer 1: $X_2$ is "H" with probability $p_2$. And the expected value
  of $p_2$ is $\frac{a_{HH}+a_{TH}}{a_{HH}+a_{HT}+a_{TH}+a_{TT}}$. Let
  $q$ denote this expected value. The posterior is 
  $$(p_{HH},p_{HT},p_{TH},p_{TT})\sim qDir(a_{HH}+1,a_{HT},a_{TH},a_{TT})+(1-q)Dir(a_{HH},a_{HT}+1,a_{TH},a_{TT})$$
  beacause $(HH)$ happens with probability $q$ and $(HT)$ occurs with
  the other probability.

Answer 2: Maybe I can't use the expected value to marginalize the missing information. So, let $p^1\sim Dir(a_{HH}+1,a_{HT},a_{TH},a_{TT})$ and $p^2\sim Dir(a_{HH},a_{HT}+1,a_{TH},a_{TT})$. Then, $$p =_D p_{HH}p^1+p_{HT}p^2,$$
where $=_D$ denotes the equality in distribution.

Answer 3: $X_1$ is head with probability $p_{HH}+p_{HT}$, so by the
  Bayes rule, 
\begin{eqnarray*} f(p|X_1=H)&\propto&
 f(X_1=H|p)f(p)\\&=&(p_{HH}+p_{HT})f(p). \end{eqnarray*}

Obviously, the two answers lead me to different posteriors.. Could anyone tell me which one should be correct? 
 A: Suppose both coins are tossed once but we only observe one of them. Assume the coin we get to observe is independent of the outcome of the toss.
Let $X_1$ and $X_2$ be random variables representing the outcome for each coin. And, let $\theta = [p_{HH}, p_{HT}, p_{TH}, p_{TT}]$ be a parameter vector indicating the joint probability of each possible pair of outcomes:
$$p(x_1, x_2 \mid \theta) = \left\{ \begin{array}{cl}
  p_{HH} & x_1=H, x_2=H \\
  p_{HT} & x_1=H, x_2=T \\
  p_{TH} & x_1=T, x_2=H \\
  p_{TT} & x_1=T, x_2=T \\
\end{array} \right.$$
Suppose we've only observed the first coin. Then, the goal is to compute the posterior distribution over parameters, given the observed outcome:
$$p(\theta \mid x_1) \propto p(x_1 \mid \theta) p(\theta)$$
where $p(\theta)$ is the prior (which you've indicated is Dirichlet). $p(x_1 \mid \theta)$ is the observed data likelihood, which is obtained by marginalizing the missing outcome out of the complete data likeilhood $p(x_1, x_2 \mid \theta)$:
$$p(x_1 \mid \theta) =
\sum_{x_2} p(x_1, x_2 \mid \theta)$$
So, if $X_1=H$, then $p(x_1 \mid \theta) = p_{HH} + p_{HT}$, and the posterior would be:
$$p(\theta \mid x_1) \propto (p_{HH} + p_{HT}) p(\theta)$$
Alternatively, if $X_1=T$, then $p(x_1 \mid \theta) = p_{TH} + p_{TT}$, and the posterior would be:
$$p(\theta \mid x_1) \propto (p_{TH} + p_{TT}) p(\theta)$$
If I understand your notation correctly, this is consistent with your 'answer 3'. A similar procedure can be used to compute $p(\theta \mid x_2)$ if only the second coin is observed.
